Calculating Velocities in One-Dimensional Collisions

[JamesRV]

We just started learning Conservation of Momentum, and I feel completely lost. This is the following question I was asked:

Homework Statement

A 2.00-kg object moving at 6.00 m/s collides with a 4.00-kg object that is initially at rest. After the collision, the 2.00-kg object moves backward at 1.00 m/s. This is a one-dimensional collision, and no external forces are acting.

a) Calculate the velocity of the 4.00-kg object after the collision.

My attempt was to simply use the Momentum equation:

Homework Equations

& The attempt at a solution[/B]

m_A1v_A1 + m_B1v_B1 = m_A2v_A2 + m_B2v_B2

So:
v_B2 = (m_A1v_A1 + m_B1v_B1 - m_A2v_A2)/m_B2

Resulting in the velocity to be 3.5m/s.

I was just looking for someone to confirm that answer, because I am unsure if I did it right.

In question b) it asks:
Calculate the velocity of the centre of mass of the two objects (i) before the collision and (ii) after the collision.
What does that mean?

Thank you for helping.

 

[Tanya Sharma]

We just started learning Conservation of Momentum, and I feel completely lost. This is the following question I was asked:

A 2.00-kg object moving at 6.00 m/s collides with a 4.00-kg object that is initially at rest. After the collision, the 2.00-kg object moves backward at 1.00 m/s. This is a one-dimensional collision, and no external forces are acting.

a) Calculate the velocity of the 4.00-kg object after the collision.

My attempt was to simply use the Momentum equation:

m_A1v_A1 + m_B1v_B1 = m_A2v_A2 + m_B2v_B2

So:
v_B2 = (m_A1v_A1 + m_B1v_B1 - m_A2v_A2)/m_B2

Resulting in the velocity to be 3.5m/s.

I was just looking for someone to confirm that answer, because I am unsure if I did it right.

In question b) it asks:
Calculate the velocity of the centre of mass of the two objects (i) before the collision and (ii) after the collision.
What does that mean?

Thank you for helping.

Hi JamesRV

Welcome to PF!

The result of part a) looks right .

For part b) use V_cm = (m1v1+m2v2)/(m1+m2) .

Apply this before and after collision .What do you get ?

Please take care of the signs .

 

[JamesRV]

Thank you Tanya!

I used the equation you gave me, it makes logic sense taking the total momentum and dividing by the mass.
I got 2m/s for both cases, which seems right!

 

[Tanya Sharma]

Thank you Tanya!

I used the equation you gave me, it makes logic sense taking the total momentum and dividing by the mass.
I got 2m/s for both cases, which seems right!

Yes..Well done :thumbs:

So what does this tell you about collision ?

 

[JamesRV]

Yes..Well done :thumbs:

So what does this tell you about collision ?

Doesn't it just mean that the momentum is conserved? I thought momentum was always conserved, is it not?

 

[Tanya Sharma]

Doesn't it just mean that the momentum is conserved? I thought momentum was always conserved, is it not?

Yes... And the velocity of Center of Mass remains unaffected in a collision.

 

[JamesRV]

Okay, thanks again!